# Does the following series converge or diverge: $\displaystyle\sum_{n=1}^{\infty}\frac{n!}{n^{n}}$?

Which of the following series converge, and which diverge?

• $$\displaystyle\sum_{n=1}^{\infty}\frac{1}{n^{2}+1}$$

• $$\displaystyle\sum_{n=1}^{\infty}\frac{n!}{n^{n}}$$

• $$\displaystyle\sum_{n=1}^{\infty}\frac{1}{\sqrt{n^{2}+n}}$$

MY ATTEMPT

The first one converges due to the comparison test. Indeed, one has \begin{align*} \sum_{n=1}^{\infty}\frac{1}{n^{2}+1} \leq \sum_{n=1}^{\infty}\frac{1}{n^{2}} \end{align*} where the last series is the $$p$$-series with $$p = 2 > 1$$.

The second series does converge due to the ratio test. Indeed, one has \begin{align*} \lim_{n\to\infty}\frac{(n+1)!}{(n+1)^{n+1}}\times\frac{n^{n}}{n!} = \lim_{n\to\infty}\left(\frac{n}{n+1}\right)^{n} = \lim_{n\to\infty}\left(1 - \frac{1}{n+1}\right)^{n} = e^{-1} < 1 \end{align*}

Finally, for the third series it suffices to notice that $$n^{2} + n \leq 2n^{2}$$. Thence we conclude that it diverges. Indeed, one has \begin{align*} \sum_{n=1}^{\infty}\frac{1}{\sqrt{n^{2}+n}} \geq \frac{1}{\sqrt{2}}\sum_{n=1}^{\infty}\frac{1}{n} \longrightarrow +\infty \end{align*}

Is the wording of my solutions good? Any comments and contributions are appreciated.

• Seems legit to me. – FearfulSymmetry Jul 1 '20 at 2:05
• Yep, looks good to me. – K.defaoite Jul 1 '20 at 3:04

Most of this is correct and well-worded. Where you have used the phrase "harmonic series with $$p = 2$$", you really should say "$$p$$-series with $$p = 2$$. The harmonic series is $$\sum_{n=1}^\infty \frac{1}{n}$$, whereas $$p$$-series are series of the form $$\sum_{n=1}^\infty \frac{1}{n^p}$$.
• I would have called the comparison series used in the first one a "p-series". I reserve "harmonic series" for the special case of $p=1$. – Jason DeVito Jul 1 '20 at 2:40